7 TWO-PORT NETWORKS Electronic circuits are frequently needed for processing a given electrical signal to extract the desired information or characteristics. This includes boosting the strength of a weak signal or ®ltering out certain frequency bands and so forth. Most of these circuits can be modeled as a black box that contains a linear network comprising resistors, inductors, capacitors, and dependent sources. Thus, it may include electronic devices but not the independent sources. Further, it has four terminals, two for input and the other two for output of the signal. There may be a few more terminals to supply the bias voltage for electronic devices. However, these bias conditions are embedded in equivalent dependent sources. Hence, a large class of electronic circuits can be modeled as two-port networks. Parameters of the two-port completely describe its behavior in terms of voltage and current at each port. These parameters simplify the description of its operation when the two-port network is connected into a larger system. Figure 7.1 shows a two-port network along with appropriate voltages and currents at its terminals. Sometimes, port-1 is called the input while port-2 is the output port. The upper terminal is customarily assumed to be positive with respect to the lower one on either side. Further, currents enter the positive terminals at each port. Since 243 Figure 7.1 Two-port network. Radio-Frequency and Microwave Communication Circuits: Analysis and Design Devendra K. Misra Copyright # 2001 John Wiley & Sons, Inc. ISBNs: 0-471-41253-8 (Hardback); 0-471-22435-9 (Electronic) the linear network does not contain independent sources, the same currents leave respective negative terminals. There are several ways to characterize this network. Some of these parameters and relations among them are presented in this chapter, including impedance parameters, admittance parameters, hybrid parameters, and transmission parameters. Scattering parameters are introduced later in the chapter to characterize the high-frequency and microwave circuits. 7.1 IMPEDANCE PARAMETERS Consider the two-port network shown in Figure 7.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, voltage V 1 at port-1 can be expressed in terms of two currents as follows: V 1  Z 11 I 1  Z 12 I 2 7:1:1 Since V 1 is in volts, and I 1 and I 2 are in amperes, parameters Z 11 and Z 12 must be in ohms. Therefore, these are called the impedance parameters. Similarly, we can write V 2 in terms of I 1 and I 2 as follows: V 2  Z 21 I 1  Z 22 I 2 7:1:2 Using the matrix representation, we can write V 1 V 2 !  Z 11 Z 12 Z 21 Z 22 ! I 1 I 2 ! 7:1:3 or, VZI7:1:4 where [Z] is called the impedance matrix of two-port network. If port-2 of this network is left open then I 2 will be zero. In this condition, (7.1.1) and (7.1.2) give Z 11  V 1 I 1     I 2 0 7:1:5 and, Z 21  V 2 I 1     I 2 0 7:1:6 244 TWO-PORT NETWORKS Similarly, with a source connected at port-2 while port-1 is open circuit, we ®nd that Z 12  V 1 I 2     I 1 0 7:1:7 and, Z 22  V 2 I 2     I 1 0 7:1:8 Equations (7.1.5) through (7.1.8) de®ne the impedance parameters of a two-port network. Example 7.1: Find impedance parameters for the two-port network shown here. If I 2 is zero then V 1 and V 2 can be found from Ohm's law as 6 I 1 . Hence, from (7.1.5) and (7.1.6), Z 11  V 1 I 1     I 2 0  6I 1 I 1  6 O and, Z 21  V 2 I 1     I 2 0  6I 1 I 1  6 O Similarly, when the source is connected at port-2 and port-1 has an open circuit, we ®nd that V 2  V 1  6 I 2 Hence, from (7.1.7) and (7.1.8), Z 12  V 1 I 2     I 1 0  6I 2 I 2  6 O IMPEDANCE PARAMETERS 245 and, Z 22  V 2 I 2     I 1 0  6I 2 I 2  6 O Therefore, Z 11 Z 12 Z 21 Z 22 !  66 66 ! Example 7.2: Find impedance parameters of the two-port network shown here. As before, assume that the source is connected at port-1 while port-2 is open. In this condition, V 1  12 I 1 and V 2  0. Therefore, Z 11  V 1 I 1     I 2 0  12I 1 I 1  12 O and, Z 21  V 2 I 1     I 2 0  0 Similarly, with a source connected at port-2 while port-1 has an open circuit, we ®nd that V 2  3 I 2 and V 1  0 Hence, from (7.1.7) and (7.1.8), Z 12  V 1 I 2     I 1 0  0 and, Z 22  V 2 I 2     I 1 0  3I 2 I 2  3 O 246 TWO-PORT NETWORKS Therefore, Z 11 Z 12 Z 21 Z 22 !  12 0 03 ! Example 7.3: Find impedance parameters for the two-port network shown here. Assuming that the source is connected at port-1 while port-2 is open, we ®nd that, V 1 12  6 I 1  18 I 1 and V 2  6 I 1 Note that there is no current ¯owing through a 3-O resistor because port-2 is open. Therefore, Z 11  V 1 I 1     I 2 0  18I 1 I 1  18 O and, Z 21  V 2 I 1     I 2 0  6I 1 I 1  6 O Similarly, with a source at port-2 and port-1 open circuit, V 2 6  3 I 2  9 I 2 and V 1  6 I 2 This time, there is no current ¯owing through a 12-O resistor because port-1 is open. Hence, from (7.1.7) and (7.1.8), Z 12  V 1 I 2     I 1 0  6I 2 I 2  6 O and, Z 22  V 2 I 2     I 1 0  9I 2 I 2  9 O IMPEDANCE PARAMETERS 247 Therefore, Z 11 Z 12 Z 21 Z 22 !  18 6 69 ! An analysis of results obtained in Examples 7.1±7.3 indicates that Z 12 and Z 21 are equal for all three circuits. In fact, it is an inherent characteristic of these networks. It will hold for any reciprocal circuit. If a given circuit is symmetrical then Z 11 will be equal to Z 22 as well. Further, impedance parameters obtained in Example 7.3 are equal to the sum of the corresponding results found in Examples 7.1 and 7.2. This happens because if the circuits of these two examples are connected in series we end up with the circuit of Example 7.3. It is illustrated here. Example 7.4: Find impedance parameters for a transmission line network shown here. This circuit is symmetrical because interchanging port-1 and port-2 does not affect it. Therefore, Z 22 must be equal to Z 11 . Further, if current I at port-1 produces an open-circuit voltage V at port-2 then current I injected at port-2 will produce V at port-1. Hence, it is a reciprocal circuit. Therefore, Z 12 will be equal to Z 21 . 248 TWO-PORT NETWORKS Assume that the source is connected at port-1 while the other port is open. If V in is incident voltage at port-1 then V in e Àg` is the voltage at port-2. Since the re¯ection coef®cient of an open circuit is 1, the re¯ected voltage at this port is equal to the incident voltage. Therefore, the re¯ected voltage reaching port-1 is V in e À2g` . Hence, V 1  V in  V in e À2g` V 2  2 V in e Àg` I 1  V in Z o 1 À e 2g`  and, I 2  0 Therefore, Z 11  V 1 I 1     I 2 0  V in 1  e À2g`  V in Z o 1 À e À2g`   Z o e g`  e Àg` e g` À e Àg`  Z o tanhg`  Z o cothg` and, Z 21  V 2 I 1     I 2 0  2V in e Àg` V in Z o 1 À e À2g`   Z o 2 e g` À e Àg`  Z o sinhg` For a lossless line, g  jb and, therefore, Z 11  Z o j tanb` ÀjZ o cotb` and, Z 21  Z o j sinb` Àj Z o sinb` 7.2 ADMITTANCE PARAMETERS Consider again the two-port network shown in Figure 7.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no ADMITTANCE PARAMETERS 249 independent sources, current I 1 at port-1 can be expressed in terms of two voltages as follows: I 1  Y 11 V 1  Y 12 V 2 7:2:1 Since I 1 is in amperes, and V 1 and V 2 are in volts, parameters Y 11 and Y 12 must be in siemens. Therefore, these are called the admittance parameters. Similarly, we can write I 2 in terms of V 1 and V 2 as follows: I 2  Y 21 V 1  Y 22 V 2 7:2:2 Using the matrix representation, we can write I 1 I 2 !  Y 11 Y 12 Y 21 Y 22 ! V 1 V 2 ! 7:2:3 or, IYV7:2:4 where [Y ] is called the admittance matrix of the two-port network. If port-2 of this network has a short circuit then V 2 will be zero. In this condition, (7.2.1) and (7.2.2) give Y 11  I 1 V 1     V 2 0 7:2:5 and, Y 21  I 2 V 1     V 2 0 7:2:6 Similarly, with a source connected at port-2 and a short circuit at port-1, Y 12  I 1 V 2     V 1 0 7:2:7 and, Y 22  I 2 V 2     V 1 0 7:2:8 Equations (7.2.5) through (7.2.8) de®ne the admittance parameters of a two-port network. 250 TWO-PORT NETWORKS Example 7.5: Find admittance parameters of the circuit shown here. If V 2 is zero then I 1 is equal to 0.05 V 1 and I 2 is À0:05 V 1 . Hence, from (7.2.5) and (7.2.6), Y 11  I 1 V 1     V 2 0  0:05V 1 V 1  0:05 S and, Y 21  I 2 V 1     V 2 0  À0:05V 1 V 1 À0:05 S Similarly, with a source connected at port-2 and port-1 having a short circuit, I 2 ÀI 1  0:05 V 2 Hence, from (7.2.7) and (7.2.8), Y 12  I 1 V 2     V 1 0  À0:05 V 2 V 2 À0:05 S and, Y 22  I 2 V 2     V 1 0  0:05 V 2 V 2  0:05 S Therefore, Y 11 Y 12 Y 21 Y 22 !  0:05 À0:05 À0:05 0:05 ! Again we ®nd that Y 11 is equal to Y 22 because this circuit is symmetrical. Similarly, Y 12 is equal to Y 21 because it is reciprocal. ADMITTANCE PARAMETERS 251 Example 7.6: Find admittance parameters for the two-port network shown here. Assuming that a source is connected at port-1 while port-2 has a short circuit, we ®nd that I 1  0:10:2  0:025 0:1  0:2  0:025 V 1  0:0225 0:325 V 1 A and if voltage across 0.2 S is V N , then V N  I 1 0:2  0:025  0:0225 0:225 ? 0:325 V 1  V 1 3:25 V Therefore, I 2 À0:2V N À 0:2 3:25 V 1 A Hence, from (7.2.5) and (7.2.6), Y 11  I 1 V 1     V 2 0  0:0225 0:325  0:0692 S and, Y 21  I 2 V 1     V 2 0 À 0:2 3:25 À0:0615 S Similarly, with a source at port-2 and port-1 having a short circuit, I 2  0:20:1  0:025 0:2  0:1  0:025 V 1  0:025 0:325 V 2 A and if voltage across 0.1 S is V M , then V M  I 2 0:1  0:025  0:025 0:125 ? 0:325 V 2  2V 2 3:25 V 252 TWO-PORT NETWORKS [...]... circuit If Vin is the incident voltage at port-1 then it will appear as Vin eÀg` at port-2 Since the re¯ection coef®cient of a short circuit is equal to À1, re¯ected voltage at this port is 180 out of phase with incident voltage Therefore, the re¯ected voltage reaching port-1 is ÀVin eÀ2g` Hence, V1 ˆ Vin À Vin eÀ2g` V2 ˆ 0 I1 ˆ Vin …1 ‡ eÀ2g` † Zo and, I2 ˆ À 2Vin Àg` e Zo Therefore, Y11 Vin  …1... has a short circuit If Vin is incident voltage at port-1 then it will be Vin eÀg` at port-2 Since the re¯ection coef®cient of the short circuit is À1, re¯ected voltage at this port is 180 264 TWO-PORT NETWORKS out of phase with incident voltage Therefore, re¯ected voltage reaching port-1 is ÀVin eÀ2g` Hence, V1 ˆ Vin À Vin eÀ2g` V2 ˆ 0 I1 ˆ Vin …1 ‡ eÀ2g` † Zo and, I2 ˆ À 2Vin Àg` e Zo Therefore,... circuit while the source is still connected at port-1 If Vin is incident voltage at port-1 then Vin eÀg` is at port-2 Since the re¯ection coef®cient of an open circuit is ‡1, re¯ected voltage at this port is equal to incident voltage Therefore, the re¯ected voltage reaching port-1 is Vin eÀ2g` Hence, V1 ˆ Vin ‡ Vin eÀ2g` V2 ˆ 2 Vin eÀg` I1 ˆ Vin …1 À eÀ2g` † Zo and, I2 ˆ 0 Now, from (7.4.6) and (7.4.7),... ith port as follows: Vi ˆ Viin ‡ Viref …7:6:9† and, Ii ˆ 1 …V in À Viref † Zoi i …7:6:10† where superscripts ``in'' and ``ref'' represent the incident and re¯ected voltages, respectively Zoi is characteristic impedance at ith port Equations (7.6.9) and (7.6.10) can be solved to ®nd incident and re¯ected voltages in terms of total voltage and current at the ith port Hence, 1 Viin ˆ Vi ‡ Zoi Ii † 2 …7:6:11†... TWO-PORT NETWORKS and, 1 Viref ˆ Vi À Zoi Ii † 2 …7:6:12† Assuming both of the ports to be lossless so that Zoi is a real quantity, the average power incident at the ith port is &  in  ' 1 1 V * 1 Piin ˆ RefViin …Iiin †*g ˆ Re Viin i jV in j2 ˆ 2 2 2Zoi i Zoi …7:6:13† and average power re¯ected from the ith port is  ref  ' o 1 & * 1 n V 1 jV ref j2 Piref ˆ Re Viref …Iiref †* ˆ Re Viref i ˆ 2 2 2Zoi... de®ned in such a way that the squares of their magnitudes represent the power ¯owing in respective directions Hence, 2 3 2 3 p Viin 1 Vi ‡ Zoi Ii 1 Vi p  ai ˆ p ˆ ˆ p p ‡ Zoi Ii 2 2 2Zoi 2 2Zoi Zoi …7:6:15† 2 3 2 3 p Viref 1 Vi À Zoi Ii 1 Vi p  bi ˆ p ˆ ˆ p p À Zoi Ii 2 2 2Zoi 2 2Zoi Zoi …7:6:16† and, Therefore, units of ai and bi are p... I1 ˆ Vin …1 ‡ eÀ2g` † Zo and, I2 ˆ À 2Vin Àg` e Zo Therefore, Y11 Vin  …1 ‡ eÀ2g` †  I1  e‡g` ‡ eÀg` 1 Zo ˆ ˆ ˆ  ˆ À2g` † V1 V2 ˆ0 Vin …1 À e Zo …e‡g` À eÀg` † Zo ? tanh…g`† and, Y21 2V  À in eÀg`  I 2 1 Zo ˆÀ ˆÀ ˆ 2 ˆ Zo …e‡g` À eÀg` † Zo ? sinh…g`† V1 V2 ˆ0 Vin …1 À eÀ2g` † For a lossless line, g ˆ jb and, therefore, Y11 ˆ 1 jZo tan…b`† and Y21 ˆ À 7.3 1 1 ˆj jZo sin…b`† Zo ? sin…b`† HYBRID... through 0.1 S is I1 À IN ˆ 0:0692V1 Using the current division rule, current IM through 0.2 S is found as follows: IM ˆ 0:2 0:0692V1 ˆ 0:0615V1 A 0:2 ‡ 0:025 Hence, I2 ˆ À…IN ‡ IM † ˆ À0:1115V1 A Now, from (7.2.5) and (7.2.6), Y11 ˆ  I1   ˆ 0:1192 S V1 V2 ˆ0 and, Y21 ˆ  I2   ˆ À0:1115 S V1 V2 ˆ0 Similarly, with a source at port-2 and port-1 having a short circuit, current I2 at port-2 is & ' 0:2…0:1... re¯ected wave Input and output re¯ection coef®cients are G1 ˆ Z1 À Zo1 b1 ˆ Z1 ‡ Zo1 a1 …7:6:21† G2 ˆ Z2 À Zo2 b2 ˆ Z2 ‡ Zo2 a2 …7:6:22† and, Dividing (7.6.17) by a1 and then using (7.6.21), we ®nd   b1 Z À Zo1 a ˆ G1 ˆ 1 ˆ S11 ‡ S12 2 a1 Z1 ‡ Zo1 a1 …7:6:23† Now, dividing (7.6.18) by a2 and then combining with (7.6.19), we get   b2 a 1 a 1 À S22 GL ˆ S22 ‡ S21 1 ˆ A 1ˆ GL a2 a2 a2 S21 GL …7:6:24† From... here The source impedance is Zo Using the voltage divider formula, we can write V2 ˆ Zo V Z ‡ 2Zo S1 Hence, from (7.6.32), S21 ˆ 2V2 2Zo 2Z ‡ Z À Z Z ˆ ˆ o ˆ1À Z ‡ 2Zo Z ‡ 2Zo VS1 Z ‡ 2Zo For evaluating S12 , we connect a voltage source VS2 at port-2 while port-1 is terminated by Zo , as shown here The source impedance is Zo Using the voltage divider rule again, we ®nd that V1 ˆ Zo V Z ‡ 2Zo S2 Now, . inductors, capacitors, and dependent sources. Thus, it may include electronic devices but not the independent sources. Further, it has four terminals, two for. There may be a few more terminals to supply the bias voltage for electronic devices. However, these bias conditions are embedded in equivalent dependent sources.